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Question

Prove that nr=03r nCr=4n.

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Solution

nr=03r nCr =

30nC0+31nC1+32nC2+...+3nnCn

=nC0(1)n.30+nC1(1)n1.31+nC2(1)n2.32+...+nCn3n

=(1+3)n=4n

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